题干：

At the beginning of every day, the first person who signs in the computer room will unlock the door, and the last one who signs out will lock the door. Given the records of signing in's and out's, you are supposed to find the ones who have unlocked and locked the door on that day.

Input Specification:

Each input file contains one test case. Each case contains the records for one day. The case starts with a positive integer M, which is the total number of records, followed by M lines, each in the format:

ID_number Sign_in_time Sign_out_time

where times are given in the format HH:MM:SS, and ID_number is a string with no more than 15 characters.

Output Specification:

For each test case, output in one line the ID numbers of the persons who have unlocked and locked the door on that day. The two ID numbers must be separated by one space.

Note: It is guaranteed that the records are consistent. That is, the sign in time must be earlier than the sign out time for each person, and there are no two persons sign in or out at the same moment.

Sample Input:

3
CS301111 15:30:28 17:00:10
SC3021234 08:00:00 11:25:25
CS301133 21:45:00 21:58:40

Sample Output:

SC3021234 CS301133

题目大意：

每天早上，第一个在电脑室签名的人会打开门，最后一个在电脑室签名的人会锁上门。根据签到记录和签到记录，你应该找到当天开门和锁门的人。

解题报告：

直接模拟即可。

AC代码：

#include<cstdio>
#include<iostream>
#include<algorithm>
#include<queue>
#include<stack>
#include<map>
#include<vector>
#include<set>
#include<string>
#include<cmath>
#include<cstring>
#define FF first
#define SS second
#define ll long long
#define pb push_back
#define pm make_pair
using namespace std;
typedef pair<int,int> PII;
const int MAX = 2e5 + 5;
int n;
map<string,int> st,ed;
char s[MAX];
int h1,m1,s1,h2,m2,s2;
int main()
{cin>>n;for(int i = 1; i<=n; i++) {scanf("%s %d:%d:%d %d:%d:%d",s,&h1,&m1,&s1,&h2,&m2,&s2);st[s] = h1*3600 + m1*60+s1;ed[s] = h2*3600 + m2*60+s2;}string stname,edname;int stt=0x3f3f3f3f,edt;for(auto x : st) {if(x.SS < stt) {stt = x.SS;stname = x.FF;}}for(auto x : ed) {if(x.SS > edt) {edt = x.SS;edname = x.FF;}}cout << stname << " " << edname << endl;return 0 ;
}